Question: $ B = \left[\begin{array}{rr}-2 & 4 \\ 4 & -2 \\ 5 & 3\end{array}\right]$ $ F = \left[\begin{array}{rr}-1 & 5 \\ -2 & 2\end{array}\right]$ What is $ B F$ ?
Solution: Because $ B$ has dimensions $(3\times2)$ and $ F$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(3\times2)$ $ B F = \left[\begin{array}{rr}{-2} & {4} \\ {4} & {-2} \\ \color{gray}{5} & \color{gray}{3}\end{array}\right] \left[\begin{array}{rr}{-1} & \color{#DF0030}{5} \\ {-2} & \color{#DF0030}{2}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ B$ , with the corresponding elements in column $j$ of the second matrix, $ F$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ B$ with the first element in ${\text{column }1}$ of $ F$ , then multiply the second element in ${\text{row }1}$ of $ B$ with the second element in ${\text{column }1}$ of $ F$ , and so on. Add the products together. $ \left[\begin{array}{rr}{-2}\cdot{-1}+{4}\cdot{-2} & ? \\ ? & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ B$ with the corresponding elements in ${\text{column }1}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{-2}\cdot{-1}+{4}\cdot{-2} & ? \\ {4}\cdot{-1}+{-2}\cdot{-2} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ B$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{-2}\cdot{-1}+{4}\cdot{-2} & {-2}\cdot\color{#DF0030}{5}+{4}\cdot\color{#DF0030}{2} \\ {4}\cdot{-1}+{-2}\cdot{-2} & ? \\ ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{-2}\cdot{-1}+{4}\cdot{-2} & {-2}\cdot\color{#DF0030}{5}+{4}\cdot\color{#DF0030}{2} \\ {4}\cdot{-1}+{-2}\cdot{-2} & {4}\cdot\color{#DF0030}{5}+{-2}\cdot\color{#DF0030}{2} \\ \color{gray}{5}\cdot{-1}+\color{gray}{3}\cdot{-2} & \color{gray}{5}\cdot\color{#DF0030}{5}+\color{gray}{3}\cdot\color{#DF0030}{2}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-6 & -2 \\ 0 & 16 \\ -11 & 31\end{array}\right] $